Linear search by a pair of distinct-speed robots

Linear search by a pair of distinct-speed robots

Evangelos Bampas1, Jurek Czyzowicz∗,2, Leszek Gąsieniec†,3, DavidIlcinkas‡,4, Ralf Klasing§,4, Tomasz Kociumaka5, and DominikPająk¶,6

1 LIF, CNRS and Aix-Marseille University, France2 Département d’informatique, Université du Québec en Outaouais, Canada3 Department of Computer Science, University of Liverpool, UK4 LaBRI, CNRS and Univ. Bordeaux, France5 Institute of Informatics, University of Warsaw, Poland6 Institute of Informatics, Wrocław University of Technology, Poland

AbstractTwo mobile robots are initially placed at the same point on an infinite line. Each robot maymove on the line in either direction not exceeding its maximal speed. The robots need to finda stationary target placed at an unknown location on the line. The search is completed whenboth robots arrive at the target point. The target is discovered at the moment when either robotarrives at its position. The robot knowing the placement of the target may communicate it to theother robot. We look for the algorithm with the shortest possible search time (i.e. the worst-casetime at which both robots meet at the target) measured as a function of the target distance fromthe origin (i.e. the time required to travel directly from the starting point to the target at unitvelocity).

We consider two standard models of communication between the robots, namely wirelesscommunication and communication by meeting. In the case of communication by meeting, arobot learns about the target while sharing the same location with the robot possessing thisknowledge. We propose here an optimal search strategy for two robots including the respectivelower bound argument, for the full spectrum of their maximal speeds. This extends the mainresult from [11] referring to the exact complexity of the problem for the case when the speed ofthe slower robot is at least one third of the faster one. In addition, we consider also the wirelesscommunication model, in which a message sent by one robot is instantly received by the otherrobot, regardless of their current positions on the line. In this model, we design an optimalstrategy whenever the faster robot is at most 6 times faster than the slower one.

1998 ACM Subject Classification F.2.2 Nonnumerical Algorithms and Problems

Keywords and phrases Linear search, Mobile robots, Group search, Different speeds

Digital Object Identifier 10.4230/LIPIcs.xxx.yyy.p

∗ Partially funded by NSERC. Part of this work was done while Jurek Czyzowicz was visiting the LaBRIas a guest professor of the University of Bordeaux.

† This work is sponsored in part by the University of Liverpool initiative Networks Systems and Techno-logies NeST.

‡ This work was partially funded by the ANR project MACARON (ANR-13-JS02-002). This study hasbeen carried out in the frame of the “Investments for the future” Programme IdEx Bordeaux – CPU(ANR-10-IDEX-03-02).

§ This work was partially funded by the ANR project DISPLEXITY (ANR-11-BS02-014).¶ This work was partially funded by the National Science Centre, Poland - grant number2015/17/B/ST6/01897.

© Evangelos Bampas, Jurek Czyzowicz, Leszek Gąsieniec, David Ilcinkas, Ralf Klasing,Tomasz Kociumaka, and Dominik Pająk

;

licensed under Creative Commons License CC-BYConference title on which this volume is based on.Editors: Billy Editor and Bill Editors; pp. 1–20

Leibniz International Proceedings in InformaticsSchloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany

http://dx.doi.org/10.4230/LIPIcs.xxx.yyy.p

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2 Linear search by a pair of distinct-speed robots

1 Introduction

Searching is a well-studied problem in which mobile robots need to find a specific targetplaced at some a priori unknown location. In some cases, a team of robots is involved, tryingto coordinate their efforts in order to minimize the time. The complexity of the multi-robotsearching is usually defined as the time when the first searcher arrives at the target positionwhose location is controlled by an adversary.

In distributed computing, one of the central problems is rendezvous when two mobilerobots collaborate in order to meet in the smallest possible time. The efficiency of therendezvous strategy is expressed as the time when the last involved robot reaches the meetingpoint, and the meeting point is arbitrary, i.e., the robots may choose the most convenientone.

In the linear search problem studied in the present paper, a pair of robots has to meet atan unknown fixed target point of the environment and the time complexity of the processis determined by the arrival of the second robot. More specifically we consider two mobilerobots placed at the origin of an infinite line. Each robot has its maximal speed that itcannot exceed while moving in either direction along the line. There is a stationary target,placed at an unknown point of the line, that a robot discovers when arriving at its placement.The robot which possesses the knowledge of the target position may communicate it tothe other robot. We consider two communication models of the robots: communication bymeeting when the robots can exchange information only while being located at the sameposition, and wireless communication when the robot finding the target may instantaneouslyinform the other robot of its position. We want to schedule the movement of both robots sothat eventually each of them arrives at the target location. The cost of the schedule is thefirst time when both robots are present at the target position. We express it as a function ofthe distance between the target and the origin.

1.1 Related workNumerous papers have been written on the searching problem, studying diverse modelsinvolving stationary or mobile targets, graph or geometric terrain, known or unknownenvironment, one or many searchers, etc. (cf. [1, 3, 17, 21]). Depending on the setting, theproblem is known under the name of treasure hunting, pursuit-evasion, cops and robbers,fugitive search games, etc. Sometimes the searching robot is not looking for an individualtarget point, attempting rather to evacuate being lost in an unknown environment ordetermine its position within a known map (e.g. [12, 15]). Several of these research papersoffer exciting challenges of combinatorial or algorithmic nature (see [17]). In most papersstudying algorithmic issues, the objective is either to determine the feasibility of the search,(i.e., whether the search will succeed under all adversarial choices) or to minimize its costrepresented by the search time, assuming some given speeds of searchers (and perhapsevaders).

Most of the time searching is considered for a single robot. As one robot usually cannotmap the graph being explored (unless e.g., leaving pebbles at some nodes; see [8]), the secondsearcher makes the task feasible (cf. [9]). However, optimization of the search by the use ofmultiple robots often involves coordination issues, where the searchers need to communicatein order to synchronize their efforts and adequately split the entire task into portions assignedto individual robots (cf. [11, 14, 16, 18]). As this objective is often not easy to achieve, somemulti-robot search problems turn out to be NP-hard (e.g., see [18]).

Several papers on searching consider online algorithms (cf. [19]), where the information

E. Bampas, J. Czyzowicz, L. Gąsieniec, D. Ilcinkas, R. Klasing, T. Kociumaka, and D. Pająk 3

about the environment is acquired as the search progresses. The performance of an onlinealgorithm is measured by its competitive ratio, i.e., the worst-case ratio of its cost withrespect to the offline cost, which is the search time of the optimal algorithm with full a prioriknowledge of the environment and the target placement. Many search problems, especiallyfor geometric environments, are analyzed from this perspective, in particular when the costof the offline solution is just the distance to the target; see [3, 11, 16, 19].

The linear search problem for a single robot was introduced by Beck [6] and Bellman [7].They proposed an optimal on-line algorithm with search time 9d, where d is the distancefrom the origin to the target. This question was extended to the cow-path problem in [2],in which the searcher has more than two directions to follow, to searching in the plane [3],and numerous other variations. Bose et al. [10] recently studied a variant of these problemswhere upper and lower bounds on the distance to the target are given. On a line, withoutthis information the time 9d cannot be improved even if the search is performed by a teamof same-speed robots communicating by meeting if all robots have to reach the target [11];see also [4]. Surprisingly, time 9d can still be achieved by distinct-speed robots if the slowestrobot is at most 3 times slower than the fastest one.

The rendezvous problem has been central to distributed computing for many years. Itwas studied in various settings (cf. [22]), but even for environments as simple as a line or aring, optimal solutions are not always known. Feasibility of the rendezvous problem is oftendetermined by a symmetry breaking process, which must prevent the robots from fallinginto an infinite pattern avoiding the meeting. Searching and rendezvous may be viewed asproblems with opposite objectives. Searching is a game between a searcher, who tries tofind the target as fast as possible and the adversary, who knows the searching strategy andattempts to maximize the search time by its choice of the environment parameters, targetplacement (or its escape route), etc. Hence in searching, the two players have contradictorygoals. In rendezvous the two players collaborate, trying to quickly find one another (see [1]).Contrary to the searching problem, the rendezvous destination is not given in advance but itmay be decided by the robots.

Equivalent to our setting are evacuation problems, where a collection of mobile robotsneed to find an unknown exit in the environment and the exit must be reached by all involvedrobots. In previous research usually robots travelling at the same speed were considered (cf.[11, 12]). For other problems considering robots with distinct speeds (e.g., the patrollingproblem studied in [13, 20]), only partial results were obtained. Optimal patrolling usingmore than two robots on a ring [13], or more than three robots on a segment [20], is unknownin general and all intuitive solutions have been proved sub-optimal for some configurations ofthe speeds of the robots. Another example is the long-standing lonely runner conjecture [23],concerning k entities moving with constant speeds around a circular track of unit-length.If the speeds are pairwise different, the conjecture states that at some moment in time allrunners are located equidistantly on the cycle. The conjecture is open in general, havingbeen verified for up to 7 runners [5].

1.2 Our resultsIn this paper, we consider the linear search problem for two robots equipped with distinctmaximal speeds. For the convenience of presentation we scale their speeds so that the speedof the faster robot is 1 and the slower one is 0 < v ≤ 1.

In the model with communication by meeting, we propose an optimal strategy for anyvalue of v. And in particular our strategy works in time 1+3v

v−v2 d, for any v ≤ 13 for the target

being placed at unknown distance d from the origin. The remaining part of the spectrum


has been covered in [11] where the authors provide: an implicit (in the limit) argument forthe lower bound 9d when the robots share the maximal speed 1; and they show that thisbound can be met from above when the slower robot’s maximal speed is at least 1

3 .

In the model with wireless communication, we design a strategy achieving search time2+v+

√v2+8v

2v d. We show that this is optimal for any v ≥ 16 . Note that for v >

√17−4 ≈ 0.123

our strategy for wireless communication outperforms the optimal strategy for communicationby meeting, which shows that the feature of wireless communication is useful. On the otherhand, one can observe that this feature becomes less significant as v decreases. For v = 1, theoptimal algorithm for wireless communication is 3 times faster than the optimal algorithmfor communication by meeting whereas for v = 1

6 , it is only 1.08 times faster.

2 Preliminaries

For any algorithm A, we denote by t(A, p) the search time of algorithm A if the target islocated at point p. We define τ(A) = lim sup|p|→∞

t(A,p)|p| as the main efficiency measure of

the algorithms. Whereas all the lower bounds we derive hold for the efficiency measure τ(A),all the algorithms A we design actually satisfy the stronger property t(A, p) ≤ τ(A)|p| forevery point p ∈ R (sometimes by making infinitesimal moves as the time approaches 0). Inconsequence, our bounds are in particular directly adaptable to a setting where the targetplacement must lie at a distance at least x from the origin, where x is a fixed constant, andone measures performance of the algorithms using sup{ t(A,p)|p| : |p| > x}.

Having fixed an algorithm A for a set R of robots, each robot Γ ∈ R follows a fixedtrajectory while it is unaware of the location of the target. We use Γ(t) to denote theposition of robot Γ at time t provided that the target location is not known to the robot.Our lower bounds rely on the analysis of the progress speeds lim supt→∞

|Γ(t)|t . The largest

of these values over Γ ∈ R is called the overall progress speed. For each point p, the timeT (p) = min{t : ∃Γ∈RΓ(t) = p} is called the discovery time of p (it is the first momentwhen any robot visits p) and φ(p) denotes the set of robots which visit p at time T (p). Tosimplify notation, we will not make explicit the dependence of Γ(t), T (p), and φ(p) on thealgorithm A. Our results are primarily designed for a set R of two of robots, denoted Rand r. Their speed limits are 1 and v (v ≤ 1), respectively.

3 Communication by meeting

In this model, once a robot finds the target, it must walk to meet the other robot, and thenthe robots travel to the target. Naturally, the schedule consists of three phases: explorationphase while the target is unknown, pursuit phase where the informed robot chases after theother one in order to tell it about the target, and target phase when both robots walk tothe target location. Recall that for robots with equal speeds, one of the possible optimalsolutions consists in both robots following together a cow-path trajectory [4, 11], thus thepursuit and target phases may be nonexistent.

3.1 The upper bound

A robot following a standard cow-path trajectory visits, in order of increasing k, the pointspk := (−2)k, k ∈ Z, on alternating sides of the origin, travelling at full speed between


t

0pkpk+1 pk+2

R r

t

0pkpk+1 pk+2

R r

Figure 1 Illustration of algorithm A∗ before target detection (left), and when the target hasbeen located (right). The horizontal axis represents the line searched and the vertical axis representsthe time. The empty circle denotes the target discovery. Double and single solid lines represent thetrajectories of the faster and the slower robot, respectively. Dashed lines correspond to the overallprogress speed and dotted lines to the search time.

consecutive points pk.1 In this strategy, the robot discovers new locations after it passes pk onthe way from pk+1 to pk+2. This happens from time tk := |pk|+2

∑k+1j=−∞ |pj | = 9 ·2k = 9|pk|

to t′k+2 := |pk+2|+2∑k+1j=−∞ |pj | = 12·2k = 3|pk+2|. Consequently, the search time is bounded

from above by 9|p|.As observed by Chrobak et al. [11], this strategy generalizes to a collection of two robots

with speed limits 1 and 13 . Both robots follow the cow-path trajectory at their maximal

speed, which means that they meet in pk at time tk = 3t′k. When the faster robot R discoversthe target at a point p between pk and pk+2, it pursues the slower robot r and brings it tothe target, which turns out to be feasible within time 9|p|; see Fig. 1.

We extend this strategy to allow v < 13 as the speed limit of the slower robot r. We insist

on the two robots meeting in points pk at times tk for adjusted values pk and tk. The smallerspeed v of r allows R to travel further before going back to pk. More formally, we increasethe ratio |pk+1|/|pk| and instead of taking pk = (−2)k, we set pk = (−c)k for some c > 2. Westill make both robots visit consecutive points pk at their full speeds, and we choose c so thatthey meet in pk while r is there for the first time and R for the second time. A conditioninductively forcing the meeting at pk to be followed by a meeting in pk+1 can be expressedas 1

v |pk+1 − pk| = tk+1 − tk = |pk+1 − pk+2|+ |pk+2 − pk|, i.e., 1v (c+ 1) = 2c2 + c− 1. This

gives c = 1+v2v , which we use for our algorithm A∗.

The following theorem bounds the search time by robots using this strategy.

I Theorem 1. For the algorithm A∗ and every point p ∈ R, we have:

t(A∗, p) = 1+3vv−v2 |p| if v ≤ 1

3 , (1)

t(A∗, p) = 9|p| if 13 < v ≤ 1. (2)

Proof. First, let us show (1). Let us choose k so that the target p is located between pk and

1 Note that the sequence (pk)k∈Z is understood as prescribing infinitesimally small moves for the robot inthe two directions around the origin at the beginning of the execution (when time is in the neighborhoodof 0, i.e., at the beginning of the execution, the robot visits points pk for k in the neighborhood of −∞,hence it makes infinitesimal moves). Algorithm A∗, described below, has similar behavior. In orderto avoid this, we could start the sequence pk from any finite k (instead of −∞). This would result insmall constant additive terms appearing throughout the calculations, but the asymptotic behavior ofthe algorithm and in particular the efficiency measure τ(A) would be unaffected.


Algorithm A∗ [for two robots with communication by meeting]

1. Until the target is located, both robots visit, in order of increasing k, the points pk = (−c)kfor all k ∈ Z, where c = 1+v

2v and v = min(v, 13 ). Robot R moves with speed 1 between

consecutive points, and robot r with speed v.2. When R finds the target, it moves with speed 1 to meet and notify r.3. After the meeting, robots move together to the target at speed v.

pk+2. The meeting time in pk is tk = 1v

(|pk|+2

∑k−1j=−∞ |pj |

)= 1

v ck(

1 + 2c−1

)= 1

v ck c+1c−1 =

ck 1+3vv−v2 . Suppose that |p− pk| = δ. After meeting r in pk, robot R needs time δ to discover

the target. At that time, the distance between the robots is δ(1 + v) since they were goingin opposite directions with their maximal speeds until time tk + δ. Then, the faster robotpursues the slower one. With the speed difference of 1− v this takes δ(1+v)

1−v units of time.Next, the robots go back to the target at speed v which requires time δ(1+v)

v−v2 , i.e., 1v times

more than the pursuit. In total, the time between tk and the moment when both robotsreach the target is δ + δ(1+v)

1−v + δ(1+v)v−v2 = δ v−v

2+v+v2+1+vv−v2 = δ 1+3v

v−v2 . Since tk = |pk| 1+3vv−v2 , the

total search time is t(A, p) = (|pk|+ δ) 1+3vv−v2 = |p| 1+3v

v−v2 , as claimed.To show (2), we simply observe that, for v = 1

3 , we have 1+3vv−v2 = 9. Note that for v > 1

3 ,the searcher moving at velocity 1

3 could increase its speed to v, but no additional gain inefficiency is possible (see the lower bounds in [11, 4] and in Section 3.2). J

3.2 The lower boundWe show that the strategy from Section 3.1 is optimal, achieving the best possible boundon the search time. In fact, some results of this section are presented in order to work forcollections R of any number of robots. Consequently, in this section v denotes the slowestmaximal speed among all the robots in R, and r denotes some robot with maximal speedv. We also define τ∗ = 1+3v

v−v2 and, for any fixed algorithm A, the overall progress speedw = maxΓ∈R lim supt→∞

|Γ(t)|t . Note that the functions Γ and w depend on A, but we do

not make this relation explicit in our notation.Before we proceed with the actual lower bound, let us prove a lemma relating the search

time and the overall progress speed for any collection R of robots.

I Lemma 2. For any algorithm A and any collection R of robots with speeds not exceeding1, we have τ(A) ≥ 1+3w

w−w2 , when w ∈ (0, 1). If w = 0 or w = 1, then τ(A) cannot be boundedfrom above by any finite number.

Proof. We proceed with a proof by contradiction. That is, we suppose that τ(A) can bebounded from above if w ∈ {0, 1}, and that τ(A) < 1+3w

w−w2 if w ∈ (0, 1). In both cases, theassumption implies the existence of a finite τ such that τ(A) < τ and (w − w2) τ < 1 + 3w.The former condition yields that there exists d0 such that t(A, p) < τ |p| for |p| ≥ d0. Wewill obtain a contradiction with respect to the latter condition.

Let us fix ε > 0. Note that there exists t0 such that |Γ(t)|t ≤ w + ε for every t ≥ t0 and

every robot Γ ∈ R. Also, there exists a robot Γ and arbitrarily large time values t such that|Γ(t)|t ≥ w − ε. We fix such a robot Γ and time t, which satisfies t ≥ (τ − 1) max(d0, t0).Let p = Γ(t) and dp = |p|. Also, consider a point q at distance dq = t+dp

τ−1 from the originon the opposite side of p; see Fig. 2. Note that dq ≥ d0, so t(A, q) < τdq.


Suppose that robot Γ at time t cannot exclude the possibility that the target is located atq. Then, it must be able to reach q by the deadline, at t(A, q) < τdq, starting at time t frompoint p. The robot cannot exceed the speed limit of 1, so we conclude τ dq − t > dp + dq.However, the distance dq is defined so that τ dq − t = dp + dq, a contradiction.

Consequently, robot Γ must already know at time t that the target is not at point q. Sincerobots can only communicate by meeting and their speeds are limited by 1, this informationneeds dq + dp time to travel from q to p. In other words, some robot Γ′ must have visited qat time t′ ≤ t− dp − dq.

On the other hand, the speed limit of Γ′ is at most 1, so we have t′ ≥ dq ≥ t0. Hence, we canuse a stronger bound using progress speed: dq = Γ′(t′) ≤ t′(w + ε). Consequently, we obtaindq ≤ (w + ε)(t− dp − dq). Plugging in the definition of dq, after some term rearrangements,we get (1+w+ε)(t+dp) ≤ (t−dp)(w+ε)(τ−1). Equivalently, dp ≤ t (w+ε)(τ−2)−1

(w+ε)τ+1 . However,recall that time t was chosen so that dp ≥ (w − ε)t. Therefore, w − ε ≤ (w+ε)(τ−2)−1

(w+ε)τ+1 .

As ε > 0 can be chosen arbitrarily close to 0, we conclude that w ≤ w(τ−2)−1wτ+1 , that is

(w − w2)τ ≥ 1 + 3w. This contradicts the definition of τ . J

The following immediate corollary gives an alternative proof of the optimality of A∗ forv ≥ 1

3 . (Recall the lower bound of 9 in [11]; see also [4].)

I Corollary 3. For any algorithm A and any collection R of robots with speeds not exceeding1, we have τ(A) ≥ 9.

Proof. It suffices to observe that 1+3ww−w2 ≥ 9 for any w ∈ (0, 1). J

We continue the analysis assuming that v < 13 and w ∈ (0, 1). We provide a series of

lemmas, each imposing certain constraints on hypothetical algorithms A satisfying τ(A) < τ∗.Eventually, we deduce that some of these constraints exclude each other. Due to spacerestrictions, in the main body of the paper, we only sketch most of the proofs, referring toAppendix A for full arguments, accompanied with figures illustrating them.

I Lemma 4. If v < 13 and τ(A) < τ∗, then w < 1−v

1+3v .

1τ ≈ w≈ w

Γ

p q

1

1

0

t′

t

τdq

dp dq

Figure 2 Illustration of notions used in the proof of Lemma 2. Rays starting from the origin aswell as thick lines representing constraints are all annotated with the corresponding speeds. Here,robot Γ, while in p at time t, must know that the target is not in q, or it must be able to reach qbefore the deadline.


Proof. Suppose that w ≥ 1−v1+3v . Note that w ≥ 1−v

1+3v >1− 1

31+1 = 1

3 (because v < 13 ) and

the function f(x) = 1+3xx−x2 is increasing on ( 1

3 , 1). Thus 1+3vv−v2 = f( 1−v

1+3v ) ≤ f(w) = 1+3ww−w2 .

Consequently, Lemma 2 implies τ(A) ≥ f(w) ≥ 1+3vv−v2 = τ∗. J

I Lemma 5. If v < 13 and τ(A) < τ∗, then lim supt→∞

|r(t)|t < vwτ∗−wv−v−w

vwτ∗+1 .

1τ ≈ wdp

t

r

p q

1

v

0

t′

t

τdq

dp dq

Figure 3 Illustration of notions used in the proof of Lemma 5. The slowest robot r, while in p attime t, must know that the target is not in q or it must be able to reach q before the deadline.

Proof sketch. We choose an arbitrarily large time t. Let p = r(t) and dp = |p|. We alsoconsider a point q at distance dq = vt+dp

vτ−1 from the origin on the opposite side of p. Here, τis an arbitrary value such that τ(A) < τ < τ∗ and 1

v < τ < τ∗ = 1v + 4

1−v . We may assumet(A, q) < τdq if t is chosen sufficiently large.

The distance dq is defined so that 1v (dq + dp) + t = τ dq. Hence, it is impossible for the

slower robot r to reach point q before τ dq > t(A, q), starting from p at time t. Consequentlyr already knows at time t that the target is not located at q. Hence, some robot must havevisited q at time t′ ≤ t− dp − dq, where the inequality is due to the fact that informationcannot travel faster than at speed 1. On the other hand, the progress speed w gives an upperbound on dq

t′ as t′ approaches infinity. We combine these two inequalities to bound dp

t fromabove and derive the claimed result. J

I Corollary 6. If v < 13 and τ(A) < τ∗, then lim supt→∞

|r(t)|t < 1

τ∗ . In particular, the set{p : r ∈ φ(p)} of points discovered by r is bounded.

Proof sketch. By Lemma 4, we may assume w < 1−v1+3v . Upon substituting this inequality

into the upper bound of Lemma 5, this implies vwτ∗−wv−v−wvwτ∗+1 < 1

τ∗ . Thus, the slowest robotvisits sufficienly far points only after the deadline. To arrive at some location earlier, it mustbe notified by some other robot about the target location. J

While Lemmas 4 and 5 and Corollary 6 hold for arbitrary collections of robots, this isnot the case for the following lemma.


|r(t)|t ≥ vW+v+W−vWτ∗

vWτ∗+1 whereW = w−w2

1+3w .


Proof sketch. By Corollary 6, we may assume that the faster robot discovers all sufficientlyfar locations. Thus, its own progress speed is equal to the overall progress speed w. Moreover,the trajectory of R can be interpreted as a search algorithm for a collection R = {R}consisting of the faster robot R only. The search time of this algorithm is T (p), and thereforeLemma 2 lets us conclude that lim sup|p|→∞

T (p)|p| ≥

1+3ww−w2 = 1

W .We choose a sufficiently far point p such that T (p)

|p| is arbitrarily close to 1W and set

dp = |p|. By Corollary 6, we may assume that the slower robot does not reach p on its ownbefore the deadline. Thus, once the faster robot discovers the target located at p, its optimalstrategy is to pursue the slower robot (moving at speed 1) and then bring it to the target(moving at speed v). We define t′ and q = r(t′) as the time and location where R catchesr. The search deadline is earlier than τ∗dp, which lets us derive a lower bound on |q|t′ andconsequently bound the progress speed of the slower robot from below. J

I Lemma 8. If v < 13 and τ(A) < τ∗, then w ≥ 1−v

1+3v .

Proof sketch. We obtain vwτ∗−wv−v−wvwτ∗+1 > vW+v+W−vWτ∗

vWτ∗+1 using Lemmas 5 and 7. Since τ∗and W are defined using v and w only, this is an inequality on these two variables. It yieldsw ≥ 1−v

1+3v as shown in the full proof. J

Lemmas 4 and 8 give conflicting constraints for any algorithm A such that τ(A) < τ∗,which implies the following theorem.

I Theorem 9. For any line search algorithm A, if v < 13 , then τ(A) ≥ τ∗.

4 Wireless communication

In this model, we have only the exploration phase and the target phase. We show that, forrobots travelling at speeds with low relative difference (i.e., if v ≥ 1

6 ), in order to achieve theoptimal search time, both robots need to participate in the exploration.

1τ∗ ≈Wdq

t′

q p

Rr

1

v

0

tt′

τ∗dp

dq

dp

Figure 4 Illustration of notions used in the proof of Lemma 7. The faster robot R, havingdiscovered at time t the target located at p, must be able to catch the slower robot r and bring it tothe target before the deadline.


t

0

r R

t

0

r R

t

0Rr

Figure 5 Illustration of algorithm B∗ before target discovery (left), when the target is discoveredby r (middle), and by R (right). The horizontal axis represents the line searched and the verticalaxis represents the time. The empty circle denotes the target discovery. Double and single solid linesrepresent the trajectories of the faster and the slower robot, respectively. Dashed lines correspondto the progress speeds of the two robots and dotted lines to the search time.

Algorithm B∗ [for two robots with wireless communication]

1. Until the target is discovered, the two robots move in opposite directions. Robot r moveswith its maximal speed v and robot R with speed v′ = 1

2 (√v2 + 8v − v).

2. When either robot finds the target, it notifies the other one using wireless communicationand the other robot moves to the target using its maximal speed.

4.1 The upper bound

The optimal strategy for two robots travelling at the same speed [4] is very simple: Bothrobots explore in opposite directions at full speeds. When a robot learns that the other robothas found the target, it changes its direction towards the target.

Let us analyze the performance of this strategy for robots with distinct speeds. The totalsearch time is a sum of three terms: the time for a robot to discover the target, the time forthe other robot to go back to the origin and the time for that robot to reach the target. Weconsider two cases. First, suppose that the faster robot R discovers the target at distance dfrom the origin. Then the total search time is d+ d+ 1

vd = (2 + 1v )d. On the other hand, if

the slower robot r discovers the target, the search time is worse: 1vd+ 1

vd+ d = ( 2v + 1)d.

Intuitively, the faster robot explores too fast and it thus spends too much time goingback to the origin. Hence, we limit the exploration speed of R to v′ < 1. When it alreadyknows the target, the faster robot is still allowed to use its full speed equal to 1. Now, thetotal search times are 1

v′ d+ 1v′ d+ 1

vd = ( 2v′ + 1

v )d and 1vd+ v′

v d+ d = 1+v′+vv d, respectively.

We choose v′ to minimize the maximal of these two quantities. As they are, respectively, adecreasing and an increasing function of v′, for the optimal value v′ these terms are equal toeach other, i.e., v′ satisfies 1+v′+v

v = 2v′ + 1

v or, equivalently, v′2 + v′v = 2v.The following fact, proved in Appendix B, gives the right values of v′ and of the search

time τ∗. This lets us complete the description of the algorithm B∗ (see Fig. 5), whose analysisfollows immediately from the discussion above.

I Fact 10. For any speed v ∈ (0, 1], define τ∗ = 2+v+√v2+8v

2v and v′ =√v2+8v−v

2 . Wehave: (a) τ∗ = 1+v+v′

v , (b) τ∗ = 1v + 2

v′ , and (c) v′2 + v′v = 2v. Moreover, if v ≥ 16 , then

3v ≥ v′ ≥ 12 .

I Theorem 11. For the algorithm B∗ we have t(B∗, p) = τ∗|p| for every p ∈ R.


1τ∗ v≈ v′ ≈ v′

R

p q

1

1

0

t

t′

τ∗dq

dp dq

Figure 6 Illustration of notions used in the proof of Lemma 12. The faster robot R, while in p attime t, must know that the target is not in q or it must be able to reach q before the deadline. Forthe former, either the slower robot r must have visited q prior to t, or R must have visited q on itsown and traveled all the way to p.

4.2 The lower boundBy Theorem 11, for all points p we have t(B∗, p) = τ∗|p| and thus τ(B∗) = τ∗. We will showthat for v ≥ 1

6 no algorithm B admits a smaller value of τ(B). As in Section 3.2, we imposesome constraints on the hypothetical algorithms, two of which are going to be inconsistent.Due to space restrictions, here we present proof sketches only; full arguments are provided inAppendix C.

I Lemma 12. If v ≥ 16 and τ(B) < τ∗, then lim supt→∞

|R(t)|t ≤ v′.

Proof sketch. For a proof by contradiction we suppose that the progress speed of R exceedsv′. Then, we may choose arbitrarily large time t such that |R(t)|

t > v′. Let p = R(t) anddp = |p|. We also consider a point q at distance dq = tv from the origin on the opposite sideof p; see Fig. 6. If the time t is chosen sufficiently large, we may assume that t(B, q) < τ∗dq.

The distance dq is defined so that the robot R is unable to reach q prior to the deadlinestarting from p at time t. Thus, some robot must visit point q at time t′ < t. The speedrestriction for the slower robot is too strong for it to arrive at q early enough. Therefore,it must be the faster robot R which discovers q. Consequently, t′ must be small enoughfor R to travel from q to p during time t− t′. On the other hand, the progress speed givesa lower bound on t′ as t approaches infinity. We combine these two bounds to derive acontradiction. J

I Lemma 13. If v ≥ 16 and τ(B) < τ∗, then the set {p : r ∈ φ(p)} of points discovered by

the slower robot r is bounded.

Proof sketch. For a proof by contradiction, we suppose that there are arbitrarily far pointsdiscovered by robot r. We choose such a point p at distance dp = |p| from the origin. Wealso consider a point q at distance dq = 2dp

τv−1 from the origin on the opposite side of p. Letτ be an arbitrary value such that τ(B) < τ < τ∗ and 1

v < τ < τ∗ = 1v + 2

v′ . We may assumet(B, p) < τdp and t(B, q) < τdq if p is chosen sufficiently far.

We analyze the discovery times t = T (p) and t′ = T (q), and distinguish two casesdepending on which is smaller. If t ≤ t′, then robot r, while in p at time t, must be able to


reach q before the deadline. The distance dq is defined so that it is unable to do so if t ≥ dp

v ,and the latter inequality easily follows from the speed limit of the slower robot r.

On the other hand, if t′ ≤ t, then the robot which visits q at time t′ must be able toreach p before τ dp. This gives an upper bound on t′ ≤ τ dq − dq − dp due to the speed limits.Combined with the bound of Lemma 12 on the progress speed, this yields a contradiction ifthe initial point p is chosen sufficiently far. J

I Lemma 14. If v ≥ 16 and τ(B) < τ∗, then the set {p : r ∈ φ(p)} is unbounded.

Proof sketch. For a proof by contradiction, we suppose that the set is bounded. Then,the faster robot cannot go to infinity in one direction only, and it must pass the originat arbitrarily large moments of time. Let us fix a sufficiently large t such that R(t) = 0.Consider two points pl and pr at distance d = tv

τv−1−v on each side of the origin. Let τ bean arbitrary value such that τ(B) < τ < τ∗ and 1 + 1

v < τ < τ∗ = 1 + 1v + v′

v . If t is chosenlarge enough, we may assume that t(B, pl) < τd and t(B, pr) < τd and that both points mustbe discovered by R.

Since R(t) = 0, points pl and pr can only be discovered no later than at time t − d orno sooner than at time t+ d. The distance d is defined so that the slower robot r startingfrom any position at time t+ d is either unable to reach pl or unable to reach pr. Hence, oneof these points must be discovered (by R) at time t′ ≤ t − d. For sufficiently large t, thiscontradicts the bound of Lemma 12 on the progress speed. J

When combined, Lemmas 13 and 14 exclude any algorithm with τ(B) < τ∗.

I Theorem 15. For any line search algorithm B, if 16 ≤ v ≤ 1, then τ(B) ≥ τ∗.

5 Conclusions and open questions

Clearly, any search strategy for the communication by meeting model may also be used inthe wireless communication model. The bound of 1+3v

v−v2 obtained in Section 3.1 outperforms2+v+

√v2+8v

2v from Section 4.1 for small values of v. More precisely, an interested reader

1τ

1τv ≈ v′

r R

p q

v

1

0

t = t′

τ dp

τ dq

dp dq

Figure 7 Illustration of notions used in the proof of Lemma 13. The slower robot r, whilediscovering p at time t, must know that the target is not in q or it must be able to reach q beforethe deadline. For the former, the robot discovering q at t′ prior to t, must be able to reach p beforethe deadline.


may observe that for v ≤√

17− 4 ≈ 0.123 we have 1+3vv−v2 ≤ 2+v+

√v2+8v

2v . This immediatelyshows that our strategy from Section 4.1 is not optimal in general. We conjecture that forv ≤√

17− 4 some variation of the strategy from Section 3.1, when the faster robot is theonly one responsible for exploration, will be optimal also for the wireless communicationmodel. (Note that for general target points, it is not possible to improve the performance ofthe algorithm from Section 3.1 for wireless communication just by making the slower robotchange direction immediately once the target is discovered by the faster robot.) As bothstrategies are fundamentally different, it would also be interesting to see what happens forthe speeds

√17− 4 < v < 1

6 .The above fact may be viewed from another, perhaps more interesting perspective. Two

unit-speed robots perform linear search in 9d time when communicating by meeting and in 3dtime for the less restrictive wireless communication. Is it true that, for significantly differentrobot speeds, the wireless communication model loses its advantage over the communicationby meeting model, and the linear search takes the same time in both models?

Another possible area of research is to extend the considerations to a larger collection ofdistinct-speed robots for both communication models.

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A Full proofs for Section 3.2


|r(t)|t < vwτ∗−wv−v−w

vwτ∗+1 .

Proof. Take τ such that τ(A) < τ < τ∗ and 1v < τ < τ∗ = 1

v + 41−v . Then there exists d0

such that t(A, p) < τ |p| if |p| ≥ d0. Fix ε > 0. There exists t0 such that |R(t)|t ≤ w + ε for

every t ≥ t0. Let us choose time t so that t ≥ (τ − 1v ) max(d0, t0).

Let p = r(t) and dp = |p|. Consider a point q at distance dq = vt+dp

vτ−1 from the origin, onthe other side of the origin than point p. Note that dq ≥ vt

vτ−1 ≥ d0, so t(A, q) < τdq.Suppose that the slowest robot r does not know at time t whether the target is in q.

Then it must be able to reach q earlier than τ dq starting from p at time t. It cannot exceedthe speed limit v, so we must have (τ dq − t)v > dp + dq. However, dq is defined so that(τ dq − t)v = dp + dq, a contradiction.

Thus, the slowest robot r must know at time t that the target is not at point q. By thespeed limit, this information needs dp + dq time to travel from q to p. In other words, somerobot must have visited q at time t′ ≤ t− dp − dq.


On the other hand, we have t′ ≥ dq ≥ t0 so dq ≤ t′(w + ε). Consequently, dq ≤(w + ε)(t− dp − dq). Taking into account the definition of dq, after term rearrangements weget

(1 + w + ε)(vt+ dp) ≤ (t− dp)(w + ε)(vτ − 1).

Equivalently, |r(t)| = dp ≤ t (vτ−v−1)(w+ε)−vvτ(w+ε)+1 .

Since t can be any sufficiently large positive real number, we have

lim supt→∞

|r(t)|t ≤ (vτ−v−1)(w+ε)−v

vτ(w+ε)+1 .

By freedom to choose arbitrary ε > 0, we infer

lim supt→∞

|r(t)|t ≤ vwτ−vw−w−v

vwτ+1 < vwτ∗−vw−w−vvwτ∗+1

where the last inequality follows from monotonicity and τ < τ∗. J

I Corollary 6. If v < 13 and τ(A) < τ∗, then lim supt→∞

|r(t)|t < 1

τ∗ . In particular, the set{p : r ∈ φ(p)} of points discovered by r is bounded.

Proof. Note that

f(x) = (vτ∗−v−1)x−vvτ∗x+1 = ( 1+3v

1−v −v−1)x−v1+3v1−v x+1 = v((3+v)x−(1−v))

(1+3v)x+(1−v)

is an increasing function for x > 0, since (3 + v) · (1− v) + (1 + 3v) · (1− v) > 0. By Lemma 4we may assume w < 1−v

1+3v . Combining f(w) < f( 1−v1+3v ) with Lemma 5 gives

lim supx→∞

|r(x)|x < f(w) < f( 1−v

1+3v ) = v((3+v) 1−v1+3v−(1−v))

(1+3v) 1−v1+3v +(1−v) = v( 3+v

1+3v−1)2 = 2v−2v2

2(1+3v) = 1τ∗ .

Consequently, lim supx→∞|r(x)|x < 1

τ∗ .In particular, for any large enough |p|, if r ∈ φ(p), we would have T (p) > τ∗|p|, a

contradiction. (Recall that T (p) denotes the discovery time of p.) J



vWτ∗+1 whereW = w−w2

1+3w .

1τ ≈ wdp

t

r

p q

1

v

0

t′

t

τdq

dp dq

Figure 8 Illustration of notions used in the proof of Lemma 5. The slowest robot r, while in p attime t, must know that the target is not in q or it must be able to reach q before the deadline.


Proof. Recall that there exists d0 such that t(A, p) < τ∗|p| if |p| ≥ d0. By Corollary 6,lim supt→∞

|r(t)|t < 1

τ∗ , so there exists t0 such that |r(t)|t < 1τ∗ for every t > t0. Moreover, we

have y := sup{|p| : r ∈ φ(p)} <∞ and thus w = lim supt→∞|R(t)|t .

Note that the trajectory of R can be interpreted as a search algorithm for a collection ofR = {R} of one robot. Its search time is equal to its discovery time T (p) and by Lemma 2we can conclude that lim sup|p|→∞

T (p)|p| ≥

1+3ww−w2 = 1

W . In other words, for every ε > 0 thereexist arbitrarily far points p such that t := T (p) ≥ |p|

W+ε . We fix ε and choose p so thatdp = |p| satisfies dp ≥ max(d0, y) and dp ≥ t0v.

The slower robot has strictly less than dpτ∗ − t time to get to p from time t if the targetis located there. Since dp > t0v, the slower robot cannot visit p prior to the deadline withoutknowing that the target is at p. Thus, the faster robot must meet the slower one at sometime t′, t < t′ < dpτ

∗ and then the two robots shall go to the target with speed v.Let q = r(t′) and let dq = |q|. We have dpτ∗ − t′ > 1

v (dp − dq) and t′ − t ≥ dp − dq, i.e.,dq

t′ > v − dp(vτ∗−1)t′ and dq

t′ ≥dp+tt′ − 1. Taking an appropriate convex combination of these

two inequalities, we conclude

dq

t′

( 1dp(vτ∗−1) + 1

dp+t)≥ v

dp(vτ∗−1) −1

dp+t ,

i.e.,|r(t′)|t′ = dq

t′ ≥v(dp+t)−dp(vτ∗−1)dp+t+dp(vτ∗−1) = vdp+tv+dp−dpvτ

∗

dpvτ∗+t .

The right-hand side is an increasing function of t, since v · dpvτ∗ − (vdp + dp − dpvτ∗) =dp(v + 1)(vτ∗ − 1) > 0. Thus, we can apply t ≥ dp

W+ε to obtain

|r(t′)|t′ ≥

v(W+ε)+v+(W+ε)−v(W+ε)τ∗v(W+ε)τ∗+1 .

Since t′ > t and t can be chosen arbitrarily large, we can replace the left-hand side withlim supt′→∞

|r(t′)|t′ and, since ε can be chosen arbitrarily small, we obtain

lim supt→∞


vWτ∗+1

as claimed. J

1τ ≈Wdq

t′

q p

Rr

1

v

0

tt′

τ dp

dq

dp

Figure 9 Illustration of notions used in the proof of Lemma 7. The faster robot R, havingdiscovered at time t the target located at p, must be able to catch the slower robot r and bring it tothe target before the deadline.


I Lemma 8. If v < 13 and τ(A) < τ∗, then w ≥ 1−v

1+3v .

Proof. Combining Lemmas 5 and 7, we get

vwτ∗ − wv − v − wvwτ∗ + 1 >

vW + v +W − vWτ∗

vWτ∗ + 1 .

Expanding with the definition of τ∗ and W gives

(3 + v)w − (1− v)w(1 + 3v) + (1− v) >

1− v −W (3 + v)W (1 + 3v) + (1− v) = 1− v − 4vw + (3 + v)w2

1− v + 4w − (1 + 3v)w2 .

Note that the denominators are positive so the following inequality is equivalent.

((3 + v)w − (1− v))(1− v + 4w − (1 + 3v)w2) >(1 − v − 4vw + (3 + v)w2)(w(1 + 3v) + (1 − v)).

The corresponding polynomial has roots at w ∈ {1, 1−v1+3v ,−

1−v3+v }. Thus, the (strict)

inequality either holds in the whole interval [0, 1−v1+3v ) or in none of these points. The left-hand

side has a negative coefficient with w0 while the right-hand side has a positive one, so thisinequality is false for w = 0, and thus for every w ∈ [0, 1−v

1+3v ). Consequently, w ≥ 1−v1+3v , as

claimed. J

B Proof of Fact 10

I Fact 10. For any speed v ∈ (0, 1], define τ∗ = 2+v+√v2+8v

2v and v′ =√v2+8v−v

2 . Wehave: (a) τ∗ = 1+v+v′

v , (b) τ∗ = 1v + 2

v′ , and (c) v′2 + v′v = 2v. Moreover, if v ≥ 16 , then

3v ≥ v′ ≥ 12 .

Proof. Recall that v′ =√v2+8v−v

2 and τ∗ = 2+v+√v2+8v

2v . Consequently,

1+v+v′v = 2+2v+2v′

2v = 2+2v+√v2+8v−v

2v = 2+v+√v2+8v

2v = τ∗,

as claimed in (a). Similarly we prove (b):

1v + 2

v′ = 1v + 4√

v2+8v−v = 1v + 4(

√v2+8v+v)

(√v2+8v−v)(

√v2+8v+v) = 1

v + 4(√v2+8v+v)

8v = τ∗.

Combining these two equalities, we get (c):

1+v+v′v = 1

v + 2v′ ,

v+v′v = 2

v′ ,

vv′ + v′2 = 2v.

Moreover,v′

v =√v2+8v−v

2v = 4√v2+8v+v

andv′ = 4v√

v2+8v+v = 4√1+8/v+1

.

Consequently, v′

v and v′ are respectively a decreasing and an increasing function of v. Forv = 1

6 we have 3v = v′ = 12 , and thus 3v ≥ v′ ≥ 1

2 for every v ≥ 16 . J


1τ v≈ v′ ≈ v′

R

p q

1

1

0

t

t′

τ dq

dp dq

Figure 10 Illustration of notions used in the proof of Lemma 12. The faster robot R, while in pat time t, must know that the target is not in q or it must be able to reach q before the deadline.For the former, either the slower robot r must have visited q prior to t, or R must have visited q onits own and traveled all the way to p.

C Full proofs for Section 4.2

I Lemma 12. If v ≥ 16 and τ(B) < τ∗, then lim supt→∞

|R(t)|t ≤ v′.

Proof. Note that there exists d0 such that t(B, p) < τ∗|p| if |p| ≥ d0.Let vmax = lim supt→∞

|R(t)|t . For a proof by contradiction suppose vmax > v′. Fix ε > 0

sufficiently small (ε < vmax − v′) and note that there exists t0 such that |R(t)|t ≤ vmax + ε for

every t ≥ t0. Also, there exist arbitrarily large time values t such that |R(t)|t ≥ vmax − ε ≥ v′.

We shall choose such t which satisfies t ≥ t0v and t ≥ d0

v .Let p = R(t) and dp = |p|. Consider a point q at distance dq = tv from the origin, on the

other side of the origin than point p. Note that dq = tv ≥ d0, so t(B, q) < τ∗dq. Suppose qwas not visited by any robot prior to time t. Then the faster robot R must be able to reachq earlier than τ∗dq starting from p at time t. As it cannot exceed the speed limit, we haveτ∗dq − t > dp + dq ≥ v′t+ dq. We have dq = tv, so this implies τ∗v − 1 > v′ + v which is acontradiction since τ∗ = 1 + 1

v + v′

v by Fact 10(a).Thus, point q must have been visited by some robot at time t′ < t. First, observe that

since v is the maximal speed of the slower robot r, point q has not been visited by r. Notethat t′ ≥ dq = tv because the maximal speed of the faster robot is 1. Consequently, t′ ≥ t0and thus dq = |R(t′)| ≤ t′(vmax + ε). Additionally, t − t′ ≥ dq + dp, (i.e., t′ ≤ t − dq − dp)by the speed limit of 1 for the faster robot when moving from q at time t′ to p by time t.Combining these inequalities we get dq ≤ (t− dq − dp)(vmax + ε), and thus

tv = dq ≤ (t− dq − dp)(vmax + ε) ≤ (t− tv − t(vmax − ε))(vmax + ε),

i.e.,v ≤ (1− v − vmax + ε)(vmax + ε).

Since ε > 0 could be chosen arbitrarily small, this yields

v ≤ (1− v − vmax)vmax

Note that f(x) = (1 − v − x)x is (strictly) decreasing for x ≥ 1−v2 , in particular for

x ≥ v′ ≥ 12 ≥

1−v2 . This implies v ≤ f(vmax) < f(v′) = (1 − v − v′)v′ = v′ − 2v ≤ v since

v′ ≤ 3v and v′2 + v′v = 2v by Fact 10. This contradiction concludes the proof. J


I Lemma 13. If v ≥ 16 and τ(B) < τ∗, then the set {p : r ∈ φ(p)} of points discovered by

the slower robot r is bounded.

1τ

1τv ≈ v′

r R

p q

v

1

0

t = t′

τ dp

τ dq

dp dq

Figure 11 Illustration of notions used in the proof of Lemma 13. The slower robot r, whilediscovering p at time t, must know that the target is not in q or it must be able to reach q beforethe deadline. For the former, the robot discovering q at t′ prior to t, must be able to reach p beforethe deadline.

Proof. Let us choose τ such that τ(B) < τ < τ∗ and 1v < τ < τ∗ = 1

v + 2v′ . Then there

exists d0 such that t(B, p) < τ |p| if |p| ≥ d0. Fix ε > 0. By Lemma 12 we can assume thatlim supt→∞

|R(t)|t ≤ v′, so there exists t0 such that |R(t)|

t ≤ v′ + ε for every t ≥ t0.For a proof by contradiction suppose there exist points p such that r ∈ φ(p) and the

distances dp = |p| to the origin are arbitrarily large. We choose such a point so that dp ≥ d0and dp ≥ max(d0, t0) (τv−1)

2 .Let q be a point at distance dq = 2dp

τv−1 from the origin on the opposite side of p. Notethat dq ≥ d0 so t(B, q) < τdq. Similarly, t(B, p) < τdp.

Let t = T (p) and t′ = T (q). First, suppose that t′ ≥ t, i.e., q was discovered after p.Then the slower robot while in p at time t must be able to reach q before τ dq. It cannotexceed its maximal speed v, so we must have (τ dq − t)v > dp + dq, i.e., dq(τ v − 1) > dp + tv.By the speed limit on the slower robot we have tv ≥ dp, so dq(τ v − 1) > 2dp. However, dq isdefined so that dq(τ v − 1) = 2dp, a contradiction.

Consequently, t′ ≤ t, i.e., one of the robots must have visited q before p was discovered.Since p was not visited prior to time t′, this robot must be able to reach p before t(B, p) startingfrom q at time t′. Recall that we have t(B, p) < τdp. Thus, we must have τ dp − t′ > dp + dqso that the robot does not exceed the maximal speed 1.

On the other hand, we have t′ ≥ dq ≥ t0 and thus dq ≤ (v′ + ε)t′. Combining these twoinequalities, we get dq

v′+ε ≤ t′ < τdp− dp− dq. Expanding with the definition of dq, we obtain

2dp

(τv−1)(v′+ε) < (τ − 1)dp − 2dp

τv−1 , i.e., 2(1 + v′ + ε) < (τ − 1)(τ v − 1)(v′ + ε). By the freedomto choose arbitrarily small ε, we get 2(1 + v′) < (τ − 1)(τ v − 1)v′. The right-hand side ismonotone with respect to τ , so

2(1 + v′) ≤ (τ − 1)(τ v − 1)v′ < (τ∗ − 1)(τ∗v − 1)v′ = 1+v′v (v + v′)v′.

Consequently, 2v < (v + v′)v′ which contradicts Fact 10. J

I Lemma 14. If v ≥ 16 and τ(B) < τ∗, then the set {p : r ∈ φ(p)} is unbounded.


1τ

1τ≈ v′ ≈ v′

p` pr

vv

11

r

R

0

tt+ d

t− d

τd

d d

Figure 12 Illustration of notions used in the proof of Lemma 14. The slower robot r at time t+ d

must either be able to reach both p` and pr before the deadline, or the faster robot R must havevisited one of these points prior to time t+ d, which actually could only happen prior to time t− d.

Proof. Let us choose τ such that τ(B) < τ < τ∗ and 1 + 1v < τ < τ∗ = 1 + 1

v + v′

v . Thenthere exists d0 such that t(B, p) < τ |p| if |p| ≥ d0. Fix ε > 0. By Lemma 12 we can assumethat lim supt→∞

|R(t)|t ≤ v′, so there exists t0 such that |R(t)|

t ≤ v′ + ε for every t ≥ t0.For a proof by contradiction, suppose that y = sup{|p| : r ∈ φ(p)} < ∞. This means

that all points farther than y from the origin are discovered by the faster robot R. Thus, Rcannot go to infinity in only one direction. Hence, there exist arbitrarily large time momentst satisfying R(t) = 0. We shall take such t so that tv

τv−1−v ≥ max(d0, y, t0).Consider two points pl and pr, both at distance d = tv

τv−1−v to the left and to the rightfrom the origin, respectively. Note that d ≥ d0 so the target located at pl or pr would needto be reached earlier than τ d. First, suppose that none of these points has been discoveredprior to time t.

Consider time moment t+ d, which is the earliest moment when one of the points pl, prmay be discovered. At this time the slower robot must be able to reach both pl and pr beforethe time τ d. The distance between these points is 2d, so the robot has to traverse at leastdistance d in time strictly less than τ d− t−d. In other words, we must have v(τ d− t−d) > d,i.e., tv < d(τ v − 1− v). However, d = tv

τv−1−v , a contradiction.Therefore, at least one of the points pl, pr must have been visited before time t. Since d ≥ y,

this point must have been discovered by the faster robot, which could do that only at timet′ ≤ t−d. On the other hand, t′ ≥ d ≥ t0, so we have d = |R(t′)| ≤ (v′+ε)t′ ≤ (v′+ε)(t−d).Expanding with the definition of d, we obtain

tvτv−1−v ≤ (v′ + ε)(t− tv

τv−1−v ),

i.e.,v ≤ (v′ + ε)(τ v − 1− 2v).

Since the choice of ε was arbitrary, we get

v ≤ v′(τ v − 1− 2v) < v′(τ∗v − 1− 2v) = v′(v′ − v),

However, by Fact 10, we have v′(v′ − v) = 2(v − vv′) ≤ v. This contradiction concludes theproof. J

Linear search by a pair of distinct-speed robots

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